ANSWERS TO “TEST YOUR KNOWLEDGE” ON PAGE 34 Answers:
1. The answer is choice “b” or “disconformity.”
Unconformities define surfaces of erosion or non-deposition. They represent gaps in the geologic record. Angular unconformities generally separate underlying tilted rock layers from overlying flat-lying beds.
Nonconformities define erosional surfaces (time gaps) that separate igneous or metamorphic rocks from overlying sedi- mentary strata.
2. The answer is choice “d” or “anhydrite” (calcium sulfate).
Dolomite or dolostone [Ca,Mg(CO3)2] and limestone (CaCO3) are calcium-magnesium carbonate or calcium carbonate, respectively, whereas rock salt (NaCl) is sodium chloride.
3. The answer is choice “c” or “a three by three matrix.”
The “stress tensor” is a second order tensor. In the three by three matrix, the quantities along the principal diagonal are the normal stresses, while the remaining values indicate shear stresses.
4. The answer is choice “d” or “cassiterite” (SnO2).
Ilmenite is an ore of titanium (FeTiO3), whereas cinnabar is an ore of mercury (HgS) and wolframite is an ore of tungsten [(Fe,Mn)WO4]. By the way, if you are too young to remember “Rin Tin Tin”, we are referring to a dog; a German Shepherd and an international movie star.
5. The answer is choice “a” or [Su = C and UCS = 2C]. The proof follows:
If we substitute Ø = 0 into (1) and (2) we obtain: P1 = P3 * tan2 (45o + Ø/2) + 2C * tan (45o + Ø/2) P1 = P3 * tan2 (45o) + 2C * tan (45o) P1 = P3 + 2C
C = ½ * (P1 - P3) S = C + Pn * tan Ø S = C + Pn * tan (0o)
S = C
From (5) and (7) we write: S = C = ½ * (P1 - P3)
Su = C Equation (9) gives the value of the undrained strength (Su).
In an “unconfined compression test” the sample is loaded to failure under no lateral confinement. Thus, for saturated clays, we may input (P3) = 0 and Ø = 0 into equation (1): P1 = P3 * tan2 (45o + Ø/2) + 2C * tan (45o + Ø/2) P1 = 0 * tan2 (45o) + 2C * tan (45o) P1 = 2C
(1)
(10) (11)
Since the “unconfined compressive strength” (UCS) in the “unconfined compression test” is defined as the value of P1 at failure, then from (11) and (9):
UCS = 2C UCS = 2 Su
(12) (13)
Equations (9), (12), and (13) provide us with the answers that we seek.
It should be pointed out that, in reality, the shear strength of saturated clays is not equal to the cohesion of the Coulomb- Mohr equation, since the value of Ø is not really zero. But that’s another topic. For practical purposes, the Ø =0 concept is applicable and allows us to solve a variety of real-world problems.
34 TPG •
Oct.Nov.Dec 2017
www.aipg.org
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